Index out of Bounds Exception in for loop.

Question

Occasionaly with the code below, i get an index out of bounds error, index 1, size 1, however the for loop should ensure that it never calls the getActorLocation method if the index is the same size as the arrayList size.
Im having one of those days, and i just cant see what error i have made.
Perhaps i need coffee? lol
Cheers
James
private void checkMemoryIntegrity(){
        Actor actor = actorList.get(actorIndex);
        ArrayList<Integer> inRangeList = getActorsInMemoryRange(actor, actor.getRange());
        inRangeList.trimToSize();
        for (int i = 0; i < inRangeList.size();i++){
            if (inRangeList.size() != 0){
                actor = brainState.getActorLocation(i); //<<<<<<<<< problem line
                if (!actorList.contains(actor)){
                    brainState.actorLocations.remove(i);
public ArrayList <Integer> getActorsInMemoryRange(Actor actor, int range){
        int i = 0;
        int x = actor.getX();
        int y = actor.getY();
        ArrayList <Integer> inRangeList = new ArrayList <Integer> ();
        Actor compActor;
        while (i< brainState.actorLocations.size())
            compActor = brainState.getActorLocation(i);
            int xDist = x - compActor.getX();
            if ( (xDist >= (-1) * range) && (xDist <= range) ){
                int yDist = y - compActor.getY();
                if ( (yDist >= (-1) * range) && (yDist <= range) ){
                    inRangeList.add(i);
            i++;
        return inRangeList;
    }

Answer 1

I was thinking it might be easier to do it this way:
Iterator<Actor> i = actorLocations.iterator();
while(i.hasNext())
    if (!actorList.contains(i.next())) {
        i.remove();
}It sounds like you have an equals() method which compares the x and y locations of the actor. If not then you will have to enclose the remove in an if block which compares the actor location. Does that make sense?
Edit: contains uses the equals method.
Returns true if this list contains the specified element. More formally, returns true if and only if this list contains at least one element e such that
(o==null ? e==null : o.equals(e)).Edited by: Edward_Kimber on May 23, 2008 11:43 AM